Question

The average banana contains 0.40 g of potassium and 0.012% of naturally occurring potassium is 40k, which has a half-life of 1.248 x 10^9 years and an atomic mass of 39.964 u. What is the activity of a freshly picked bunch containing 6 bananas?

Answer 1

The activity of potassium-40 in a bunch of 6 bananas is calculated by finding the mass of 40K in the bananas and then using the decay activity formula to get approximately 75.63 Bq.

Step-by-step explanation:

The student's question is asking about the activity of potassium-40 (40K) in a bunch of bananas. To calculate the activity, we first need to find the total amount of 40K in the bananas. Since each banana contains 0.40 g of potassium and 0.012% of that potassium is 40K, we can calculate the amount of 40K in each banana, then multiply by the number of bananas, which is six. Lastly, using the half-life and atomic mass given, we can use the decay law or activity formula to find the activity in becquerels (Bq).

Assuming 1 g of 40K has an activity of 2.626 × 105 Bq, the calculation for the activity of 40K in 6 bananas would involve these steps:

1. Determine the mass of 40K in one banana: (0.40 g K/banana) × (0.012 / 100) = 4.8 × 10-5 g of 40K/banana.
2. Find the total mass of 40K in six bananas: 4.8 × 10-5 g/banana × 6 bananas = 2.88 × 10-4 g of 40K.
3. Convert the mass of 40K to kilograms (since 1 g = 1 × 10-3 kg), resulting in 2.88 × 10-7 kg of 40K.
4. Use the given activity rate for 1 g of 40K to find the activity for the mass in the bunch: (2.626 × 105 Bq/g) × (2.88 × 10-4 g) = 75.63 Bq.

Therefore, the activity of a freshly picked bunch of 6 bananas is approximately 75.63 Bq.